Let f(x)=int_0^x g(t) log _eleft(frac{1-t}{1+ | vedclass

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(C) Given $f(x)=\int_0^x g(t) \ln \left(\frac{1-t}{1+t}\right) d t$.Since $g(t)$ is an odd function and $\ln \left(\frac{1-t}{1+t}\right)$ is an odd f

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$2$

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(C) Given $f(x)=\int_0^x g(t) \ln \left(\frac{1-t}{1+t}\right) d t$.Since $g(t)$ is an odd function and $\ln \left(\frac{1-t}{1+t}\right)$ is an odd function,their product is an even function.However,the integral of an even function from $0$ to $x$ results in an odd function. Thus,$f(-x) = -f(x)$,meaning $f(x)$ is an odd function.Let $I = \int_{-\pi / 2}^{\pi / 2} \left(f(x) + \frac{x^2 \cos x}{1+e^x}\right) d x$.Using the property $\int_a^b h(x) dx = \int_a^b h(a+b-x) dx$,we have $I = \int_{-\pi / 2}^{\pi / 2} \left(f(-x) + \frac{(-x)^2 \cos(-x)}{1+e^{-x}}\right) d x$.Since $f(x)$ is odd,$f(-x) = -f(x)$. Also,$\frac{x^2 \cos x}{1+e^{-x}} = \frac{x^2 \cos x \cdot e^x}{e^x+1}$.Adding the two expressions for $I$: $2I = \int_{-\pi / 2}^{\pi / 2} \left(f(x) - f(x) + \frac{x^2 \cos x}{1+e^x} + \frac{x^2 e^x \cos x}{1+e^x}\right) d x = \int_{-\pi / 2}^{\pi / 2} x^2 \cos x d x$.Since $x^2 \cos x$ is an even function,$2I = 2 \int_0^{\pi / 2} x^2 \cos x d x$,so $I = \int_0^{\pi / 2} x^2 \cos x d x$.Using integration by parts: $I = [x^2 \sin x]_0^{\pi / 2} - \int_0^{\pi / 2} 2x \sin x d x = \frac{\pi^2}{4} - 2([-x \cos x]_0^{\pi / 2} + \int_0^{\pi / 2} \cos x d x) = \frac{\pi^2}{4} - 2(0 + [\sin x]_0^{\pi / 2}) = \frac{\pi^2}{4} - 2$.Comparing with $\left(\frac{\pi}{\alpha}\right)^2 - \alpha$,we get $\alpha = 2$.

Let $f(x)=\int_0^x g(t) \log _e\left(\frac{1-t}{1+t}\right) d t$,where $g$ is a continuous odd function. If $\int_{-\pi / 2}^{\pi / 2}\left(f(x)+\frac{x^2 \cos x}{1+e^x}\right) d x=\left(\frac{\pi}{\alpha}\right)^2-\alpha$,then $\alpha$ is equal to..............

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